Chapter 14, Problem 68P

To determine

The net head produced by the centrifugal pump in $\text{cm}$.

The required brake horse power in $\text{W}$.

Explanation of the angle at which fluid impinges on the impeller blade a critical parameter in the design of centrifugal pump.

Whether the small amount of reverse swirl is desirable or not.

Answer to Problem 68P

The net head produced by the centrifugal pump in $\text{cm}$ is $0.4649\text{cm}$.

The required brake horse power in $\text{W}$ is $21721.485\text{W}$.

The small amount of reverse swirl is not desirable.

Explanation of Solution

Given information:

The number of rotations of centrifugal pump is $750\text{rpm}$. The inlet radius is $12\text{cm}$ and the blade width is $18\text{cm}$. The outlet radius is $24\text{cm}$ and the outlet blade width is $16.2\text{cm}$. The volume flow rate is $0.573{\text{m}}^3/\text{s}$. Assuming $100\%$ efficiency. The inlet angle of blade is $-10°$ and the outlet angle of blade is $35°$.

Expression for the angular velocity for the pump

  $\omega =\frac{2\pi \dot{n}}{60}$  .......(I)

Here, the angular velocity of the centrifugal pump is $\omega$ and the number of rotation is $\dot{n}$.

Expression for the normal velocity component at the outlet of the pump

  $\text{}{V}_2=\frac{\dot{V}}{2\pi r_2{b}_2}$  .......(II)

Here, the volume flow rate is $\dot{V}$, the velocity at the outlet of the pump is $V_2$, the outlet radius is $r_2$ and the blade width at the outlet is $b_2$.

Expression for the tangential velocity component at the outlet of the pump

  $\text{}{V}^{\prime }=V_2\tan {\alpha }_2$  .......(III)

Here, the angle at the outlet of the pump is ${\alpha }_2$ and the tangential velocity component at the outlet of the pump is $V^{\prime }$.

Expression for the normal velocity component at the inlet of the pump

  $\text{}{V}_1=\frac{\dot{V}}{2\pi r_1{b}_1}$  .......(IV)

Here, the volume flow rate is $\dot{V}$, the velocity at the inlet of the pump is $V_1$, the inlet radius is $r_1$ and the blade width at the inlet is $b_1$.

Expression for the tangential velocity component at the inlet of the pump

  $\text{}{V^{\prime }}^{\prime }=V_1\tan {\alpha }_1$  .......(V)

Here, the angle at the inlet of the pump is ${\alpha }_1$ and the tangential velocity component at the inlet of the pump is ${V^{\prime }}^{\prime }$.

Expression for the equivalent head

  $\text{}{H}_{\text{water}}=H\left(\frac{{\rho }_{\text{air}}}{{\rho }_{\text{water}}}\right)$  .......(VI)

Here, the equivalent head is $H_{\text{water}}$, the density of air is ${\rho }_{\text{air}}$, the density of water is ${\rho }_{\text{water}}$ and the acceleration due to gravity is $g$.

Expression for the horse power

  $P={\rho }_{\text{water}}g{H}_{\text{water}}\dot{V}$  .......(VII)

Here, the brake horse power is $P$.

Expression for the net head

  $\text{}H=\left(\frac{\omega }{g}\right)\left(r_2{V}^{\prime }-r_1{V}^{″}\right)$   ....... (VIII)

Here, the net head is $H$.

Calculation:

Substitute $750\text{rpm}$ for $\dot{n}$ in Equation (I).

  $\begin{array}{c}\omega =\frac{2\pi \left(750\text{rpm}\right)}{60}\\ =\frac{4712.38898\text{rpm}\left(\frac{\text{1}\text{rad}/\text{s}}{\text{1}\text{rpm}}\right)}{60}\\ =78.5398\text{rad}/\text{s}\end{array}$

Substitute $0.573{\text{m}}^3/\text{s}$ for $\dot{V}$, $24\text{cm}$ for $r_2$ and $16.2\text{cm}$ for $b_2$ in Equation (II).

  $\begin{array}{c}\text{}{V}_2=\frac{0.573{\text{m}}^3/\text{s}}{2\pi \left(24\text{cm}\right)\left(16.2\text{cm}\right)}\\ =\frac{0.573{\text{m}}^3/\text{s}}{2\pi \left(24\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\left(16.2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\\ =2.34557\text{m}/\text{s}\end{array}$

Substitute $35°$ for ${\alpha }_2$ and $2.34557\text{m}/\text{s}$ for $V_2$ in Equation (III).

  $\begin{array}{c}\text{}{V}^{\prime }=\left(2.34557\text{m}/\text{s}\right)\tan \left(35°\right)\\ =\left(2.34557\text{m}/\text{s}\right)\left(0.70002\right)\\ =1.64238\text{m}/\text{s}\end{array}$

Substitute $0.573{\text{m}}^3/\text{s}$ for $\dot{V}$, $12\text{cm}$ for $r_1$ and $18\text{cm}$ for $b_1$ in Equation (IV).

  $\begin{array}{c}\text{}{V}_1=\frac{0.573{\text{m}}^3/\text{s}}{2\pi \left(12\text{cm}\right)\left(18\text{cm}\right)}\\ =\frac{0.573{\text{m}}^3/\text{s}}{2\pi \left(12\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\left(18\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\\ =4.222\text{m}/\text{s}\end{array}$

Substitute $-10°$ for ${\alpha }_1$ and $4.222\text{m}/\text{s}$ for $V_1$ in Equation (V).

  $\begin{array}{c}\text{}{V^{\prime }}^{\prime }=\left(4.222\text{m}/\text{s}\right)\tan \left(-10°\right)\\ =\left(4.222\text{m}/\text{s}\right)\left(-0.176326\right)\\ =-0.7444\text{m}/\text{s}\end{array}$

Substitute $24\text{cm}$ for $r_2$, $12\text{cm}$ for $r_1$, $78.5398\text{rad}/\text{s}$ for $\omega$, $998\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{water}}$, $1.2\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{air}}$, $-0.7444\text{m}/\text{s}$ for ${V^{\prime }}^{\prime }$, $1.64238\text{m}/\text{s}$ for $V^{\prime }$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in Equation (VI).

  $\begin{array}{c}\text{}{H}_{\text{water}}=\left[\left(\frac{78.5398\text{rad}/\text{s}}{9.81\text{m}/{\text{s}}^2}\right)\left(\begin{array}{l}\left(24\text{cm}\right)\left(1.64238\text{m}/\text{s}\right)\\ -\left(12\text{cm}\right)\left(-0.7444\text{m}/\text{s}\right)\end{array}\right)\left(\frac{1.2\text{kg}/{\text{m}}^3}{998\text{kg}/{\text{m}}^3}\right)\right]\\ =\left[\left(8\text{s}/\text{m}\right)\left(\begin{array}{l}\left(24\text{cm}\times \left(1.64238\text{m}/\text{s}\right)\right)\\ +\left(12\text{cm}\times \left(0.7444\text{m}/\text{s}\right)\right)\end{array}\right)\left(0.0012024\right)\right]\\ =0.4649\text{cm}\end{array}$

Substitute $24\text{cm}$ for $r_2$, $12\text{cm}$ for $r_1$, $78.5398\text{rad}/\text{s}$ for $\omega$, $-0.7444\text{m}/\text{s}$ for ${V^{\prime }}^{\prime }$, $1.64238\text{m}/\text{s}$ for $V^{\prime }$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in Equation (VIII).

  $\begin{array}{c}\text{}H=\left[\left(\frac{78.5398\text{rad}/\text{s}}{9.81\text{m}/{\text{s}}^2}\right)\left(\begin{array}{l}\left(24\text{cm}\right)\left(1.64238\text{m}/\text{s}\right)\\ -\left(12\text{cm}\right)\left(-0.7444\text{m}/\text{s}\right)\end{array}\right)\right]\\ =\left[\left(8\text{s}/\text{m}\right)\left(\begin{array}{l}\left(24\text{cm}\times \left(1.64238\text{m}/\text{s}\right)\right)\\ -\left(12\text{cm}\times \left(0.7444\text{m}/\text{s}\right)\right)\end{array}\right)\right]\\ =387.2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =3.872\text{m}\end{array}$

Substitute $3.872\text{m}$ for $H$, $9.81\text{m}/{\text{s}}^2$ for $g$, $998\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{water}}$, $0.573{\text{m}}^3/\text{s}$ for $\dot{V}$ in Equation (VII).

  $\begin{array}{c}P=\left(998\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(3.872\text{m}\right)\left(0.573{\text{m}}^3/\text{s}\right)\\ =\left(998\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(3.872\text{m}\right)\left(0.573{\text{m}}^3/\text{s}\right)\\ =21721.485\text{kg}\cdot {\text{m}}^2/{\text{s}}^3\left(\frac{1\text{W}}{1\text{kg}\cdot {\text{m}}^2/{\text{s}}^3}\right)\\ =21721.485\text{W}\end{array}$

Substitute $0°$ for ${\alpha }_1$ and $4.222\text{m}/\text{s}$ for $V_1$ in Equation (V).

  $\begin{array}{c}\text{}{V^{\prime }}^{\prime }=\left(4.222\text{m}/\text{s}\right)\tan \left(0°\right)\\ =\left(4.222\text{m}/\text{s}\right)\left(0\right)\\ =0\end{array}$

Substitute $24\text{cm}$ for $r_2$, $12\text{cm}$ for $r_1$, $78.5398\text{rad}/\text{s}$ for $\omega$, $998\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{water}}$, $1.2\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{air}}$, $0$ for ${V^{\prime }}^{\prime }$, $1.64238\text{m}/\text{s}$ for $V^{\prime }$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in Equation (VI).

  $\begin{array}{c}\text{}{H}_{\text{water}}=\left[\left(\frac{78.5398\text{rad}/\text{s}}{9.81\text{m}/{\text{s}}^2}\right)\left(\begin{array}{l}\left(24\text{cm}\right)\left(1.64238\text{m}/\text{s}\right)\\ -\left(12\text{cm}\right)\left(0\right)\end{array}\right)\left(\frac{1.2\text{kg}/{\text{m}}^3}{998\text{kg}/{\text{m}}^3}\right)\right]\\ =\left(8\text{s}/\text{m}\right)\left(24\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\left(1.64238\text{m}/\text{s}\right)\left(0.0012024\right)\\ =3.75\times {10}^{-3}\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ =0.375\text{cm}\end{array}$

Substitute $24\text{cm}$ for $r_2$, $12\text{cm}$ for $r_1$, $78.5398\text{rad}/\text{s}$ for $\omega$, $0$ for ${V^{\prime }}^{\prime }$, $1.64238\text{m}/\text{s}$ for $V^{\prime }$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in Equation (VIII).

  $\begin{array}{c}\text{}H=\left[\left(\frac{78.5398\text{rad}/\text{s}}{9.81\text{m}/{\text{s}}^2}\right)\left(\begin{array}{l}\left(24\text{cm}\right)\left(1.64238\text{m}/\text{s}\right)\\ -\left(12\text{cm}\right)\left(0\right)\end{array}\right)\right]\\ =\left(8\text{s}/\text{m}\right)\left(24\text{cm}\right)\left(1.64238\text{m}/\text{s}\right)\\ =315.49\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =3.1549\text{m}\end{array}$

Substitute $3.1549\text{m}$ for $H$, $9.81\text{m}/{\text{s}}^2$ for $g$, $998\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{water}}$ AND $0.573{\text{m}}^3/\text{s}$ for $\dot{V}$ in Equation (VII).

  $\begin{array}{c}P=\left(998\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(3.1549\text{m}\right)\left(0.573{\text{m}}^3/\text{s}\right)\\ =17698.634\text{kg}\cdot {\text{m}}^2/{\text{s}}^3\left(\frac{1\text{W}}{1\text{kg}\cdot {\text{m}}^2/{\text{s}}^3}\right)\\ =17698.634\text{W}\end{array}$

If the angle is $0°$, then the horse power is $17698.634\text{W}$ and net positive head is $0.375\text{cm}$ but if the angle is $-10°$, then the horse power is $21721.485\text{W}$ and the net positive head is $0.4649\text{cm}$. There is a swirl at the inlet of the pump, the net head produced by the pump and required brake horse power is decreased. The angle at which the fluid impinges has an important role in the design of centrifugal pump.

The head of the pump increases to small extent, there is a large increase in the required brake horse power. The small amount of reverse swirl is not desirable.

Conclusion:

The net head produced by the centrifugal pump in $\text{cm}$ is $0.4649\text{cm}$.

The required brake horse power in $\text{W}$ is $21721.485\text{W}$.

The small amount of reverse swirl is not desirable.